Question

An ethylene glycol solution contains 23.2 g of ethylene glycol (C2H6O2) in 98.6 mL of water. (Assume a density of 1.00 g/mL for water.) Determine freezing point and boiling point

Answer 1

Volujme of water (solvent)= 98.6ml*1 g/ml= 98.6 gm =0.0986 kg

Molecular weight of ethylene glycol= 23.2 gms=23.2gms

molecular weight of ethylene glycol (C2H6O2)=2*12+6+32=62

Moles of ethylene glycol solute = mass/ molecular weight= 23.2/62=0.3742 moles

Molality= moles of solute/ kg of solvent =0.3742/0.0986=3.8 moles of solute/kg of solvent

for boiling point

from delT= kb*m where kb= boiling point elevation constant =0.512deg.c/m=0.512 deg.c/m

del T= 0.512*3.8 =1.95, boiling point = 100+1.95= 101.95 deg.c

for freezingpoint

delT= Kf*m , kf= freezing point depression constant =1.86 deg.c/m

delT=1.86*3.86= 7.2

Freezing point =0-7.2 =7.2 deg.c